Recent content by bobinthebox

@Chestermiller Thanks a lot, ##u_1 =0## at that end is finally clear now. For what concerns the ##e_2 \cdot S(-e_1)=e_3 \cdot S(-e_1)=0## conditions, what happens if I drop them and I leave ##u_1=0## there? I know they means no shear stress on that end face, but I don't see physically what...

@Chestermiller Sorry, I really misunderstood your last question. Yes, that is precisely what I got, by exploiting symmetry. I got your ##\lambda## to be ##\frac{b+\delta}{b}##, which is equivalent to yours (##b=1##).What is striking me is that to constraint the face with normal ##-e_1## to be...

@Chestermiller For a Neo-Hookean material, I have that Cauchy is ##T = - \pi I + \mu B## where ##B=F F^t## and ##F## is the deformation gradient. Hence first Piola-Kirchoff stress is ##S=\mu F - \pi F^{-t}##. How does those relations imply the presence of a normal force on the face? Thinking...

I reflected a bit on your last useful message @Chestermiller. It seems to me that ##u_1=0## on the surface with normal ##-e_1##, which constrains that face to do not move, implies ##e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)= e_1 \cdot S(-e_1)=0##. So first I notice that the conditon is ##u_1=0## on...

Oh yes, now I see. Indeed Poisson's ratio quantifies the "amount of shrinking". This image also helps Therefore on my block there's only ##S_{11}## which is acting, all the other components of PK stress are ##0##. To conclude this thread, I just need another check: if I want to constrain a face...

That's a tricky point for me: you're saying that ##S_{12}=0##, i.e. shearing on lateral faces is 0. But during the deformation those faces are "shrunk", how is it possible that there's no stress on them? I think this is the crucial point I am missing. @Chestermiller

I got confused, you're right. So, with the conditions I imposed after your first message, the block will be stretched along direction ##e_1##, as you said. But this implies a displacement along ##e_2##, because the two lateral surfaces will be shrunk a little bit, right? I think this has to...

I've never done a FEA, I'm still math student. So, ##u_2=0## on the lateral surfaces makes perfectly sense: no displacement along the ##e_2## direction on the lateral faces. And with ##u_3=0## I have correctly no vertical displacement. Given those conditions only, I still can't determine the...

First of all, thanks as usual @Chestermiller for your answer. That's what I thought: ##S_{22}=0## implies that I do not have traction on the lateral faces. And then, also ##e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)=0## implies no shearing on the face with normal ##-e_1##. And ##e_2 \cdot S(e_1) =...

[Mentor Note -- Thread moved to the ME forum to get better views] Let's consider an incompressible block of Neo-Hookean material. Let the initial reference geometry be described by ##B=[0,b] \times [0,b] \times [0,h]##. The professor gave me the following task: Of course there can be many...

Thanks so much @Chestermiller. For what concerns my "formal" proof on the consequence for frame indifference of taking only the velocity gradient rather than its symmetric part is okay, right?

Thanks to both of you guys for your helpful answers. @Chestermiller So do you agree with my "formal" argument? I've not been able to find this on the web, nor in the books I have. I can't understand why you say "one would predict a state of stress that depends on the rotation rate of the...

@caz Could you be more explicit? I'm really sorry but I can't understand what you mean with

During lecture today, we were given the constitutive equation for the Newtonian fluids, i.e. ##T= - \pi I + 2 \mu D## where ##D=\frac{L + L^T}{2}## is the symmetric part of the velocity gradient ##L##. Dimensionally speaking, this makes sense to me: indeed the units are the one of a pressure...

Yes, because you used the fact that ##T(-n) = -T(n)##, where now ##n = e_r##. Since ##T e_r = T_{rr} e_r + T_{r \theta} e_{\theta}## and the normal at ##r= r_i## is ##-e_r## we have ## T(-e_r)=-T(e_r)=-T_{rr} e_r - T_{r \theta} e_{\theta}##, as you wrote. I feel like I got it...